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madreJ [45]
3 years ago
10

A coffee distributor needs to mix a(n) House coffee blend that normally sells for $10.50 per pound with a Kenya coffee blend tha

t normally sells for $12.80 per pound to create 40 pounds of a coffee that can sell for $10.67 per pound. How many pounds of each kind of coffee should they mix?
Answer: They must mix
______pounds of the House Blend
______pounds of the Kenya Blend.
Round your answers to the nearest whole number of pounds.
Mathematics
1 answer:
Leto [7]3 years ago
7 0

Answer:

They must mix

37 pounds of the House Blend

3 pounds of the Kenya Blend.

Step-by-step explanation:

Let 'x' be House coffe and 'y' be Kenya Coffee. Then:

10.50x + 12.80y = 10.67(40)

Also, we knw that:

x + y = 40

Solving the system of equations above, we have that:

x ≈37.0435 ≈ 37 pounds

y ≈2.95652 ≈ 3 pounds

They must mix

37 pounds of the House Blend

3 pounds of the Kenya Blend.

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Volgvan

Answer:

the answer should be 9

Step-by-step explanation:

(256 * 3^{-5} * 6^{0} )^{-2} * (\frac{3^{-2}}{2^{3}} )^{4} * 2^{28}

(256*\frac{1}{3^{5}} *6^{0})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}

(256*\frac{1}{243} *6^{0})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}

(\frac{256}{243} *6^{0})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}

(\frac{256}{243})^{-2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}

(\frac{243}{256})^{2}*(\frac{3^{-2}}{2^{3}})^{4}*2^{28}

\frac{59049}{65536}*(\frac{1}{2^{3}*3^{2}})^{4}*2^{28}

\frac{59049}{65536}*(\frac{1}{8*9})^{4}*2^{28}

\frac{59049}{65536}*(\frac{1}{72})^{4}*2^{28}

\frac{59049}{65536}*\frac{1}{26873856}*2^{28}

\frac{9}{65536}*\frac{1}{4096}*2^{28}

\frac{9}{65536*4096}*2^{28}

\frac{9}{268435456}*268435456

9

6 0
3 years ago
Read 2 more answers
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balandron [24]
3. 52h
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castortr0y [4]
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Shaun is picking out some movies to rent, and he is primarily interested in documentaries and mysteries. He has narrowed down hi
ycow [4]

Answer:

The total number of combinations in which the movies can be rented such that there is at least one documentary is 27589.

Step-by-step explanation:

The total movies are given as 19 mysteries +12 documentaries=19+12=31

Now the number of total combinations such that 4 movies are rented is given as ^{31}C_4=31465

Now the number of total combinations with no documentary is given as

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So the number of combination such that at least one movie is documentary is given as

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So the total number of combinations in which the movies can be rented such that there is at least one documentary is 27589.

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4 years ago
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LuckyWell [14K]

Answer:

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Step-by-step explanation:

all you have to do is divide the two bottom numbers on each problem and the answer to the problem will be the same on top and bottom

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4 years ago
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